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If the following two assumptions are provided:

A. $\displaystyle \lim_{x \to a^+} f'(x)=L$

B. $f$ is continuous at $a$,

could someone please explain how exactly this resulting statement is constructed: $$\lim_{x \to a^+}f'(x)=\lim_{x \to a^+}f'(\eta_x)$$, where $\eta_x$ is described as the value in $(a,x)$ that is provided by the Mean Value Theorem with the following property: $f'(\eta_x)=\frac{f(x)-f(a)}{x-a}$.

I have an attempt at explaining this claim, but I believe it requires the axiom of choice...so I am unsure if it is correct / the best way of formally explaining this result.

Attempted Explanation

  1. The assumption that $\displaystyle \lim_{x \to a^+}f'(x)=L$ implicitly tells us that there is some $b \gt a$ such that $f'$ is defined on the interval $(a,b]$.

  2. The assumption that $f$ is continuous at $a$ let's us invoke the Mean Value Theorem on the interval $[a,b]$ and any subinterval of the form $[a,x]$.

  3. By the Mean Value Theorem, we know that for any $x \in (a,b]$, there is an $\eta$ such that $f'(\eta)=\frac{f(x)-f(a)}{x-a}$ and $\eta \in (a,x)$

  4. From 3., we can define the following function $M$ whose domain is $(a,b]$ and whose mapping rule is: $M(x)=\eta$ such that $f'(\eta)=\frac{f(x)-f(a)}{x-a}$ and $\eta \in (a,x)$. $\color{red}{\text{NOTE}}$ - I believe this function definition requires one to rely on the Axiom of Choice, because it may be the case that there are an infinite number of different values that can satisfy the relevant properties for a given $x$ input. The easiest example I can think of is a linear equation.

  5. With this definition, we can now prove that $\displaystyle \lim_{x \to a^+}f'(x)=L \rightarrow \lim_{x \to a^+}f'\circ M(x)=L$ (which is quick and painless)

Consider an arbitrary $\varepsilon \gt 0$ and the corresponding $\delta_{\varepsilon}$ given to us by assumption. We know that for any $x \in (a,a+\delta_{\varepsilon}):|f'(x)-L|\lt \varepsilon$. Choose any $x \in (a,a+\delta_{\varepsilon})$. Let $\delta_1=x-a$. Then, for all $z \in (a,a+\delta_1):M(z) \in (a,a+\delta_1) \subset (a,a+\delta_{\varepsilon})$, which implies that for any $z \in (a,a+\delta_1): |f'\left(M(z)\right)-L| \lt \varepsilon$.

This is what is meant by $\displaystyle \lim_{x \to a^+}f'(x)=\lim_{x \to a^+}f'(\eta_x)$.

Edit: This post is meant to specifically look at the equality $\displaystyle \lim_{x \to a^+}f'(x)=\lim_{x \to a^+}f'(\eta_x)$...as it is through this equality that we arrive at the conclusion: $\displaystyle \lim_{x \to a^+}f'(x)=f'^+(a)$ because $\lim_{x \to a^+}f'(\eta_x)=\lim_{x \to a^+}\frac{f(x)-f(a)}{x-a}:=f'^+(a)$

S.C.
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    That claim is not true unless we assume something, for example it's ok if $f'$ is continuous... – David C. Ullrich Aug 23 '22 at 08:30
  • @DavidC.Ullrich could you please explain why we require that $f'$ is continuous? I cannot determine why this is necessary (in particular, it is not clear to me why the little proof following point 5 is wrong) – S.C. Aug 23 '22 at 10:21
  • Since $f'(\eta_x)=\frac{f(x)-f(a)}{x-a}$ it follows from the definition of the derivative that $\lim_{x\to a}f'(\eta_x)=f'(a)$. But in general $f'(a)\ne\lim_{x\to a}f'(x)$. – David C. Ullrich Aug 23 '22 at 10:23
  • @DavidC.Ullrich Maybe I am overlooking something, but the fact that I assumed that $f$ is continuous at $a$ necessarily implies that $f'^+(a)=\lim_{x \to a^+} f'(x)$...doesn't it? – S.C. Aug 23 '22 at 10:30
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    No, it certainly does not imply that. You're basically saying that if $f$ is continuous then $f'$ is continuous, simply not true – David C. Ullrich Aug 23 '22 at 10:31
  • Let $f(x)=x^2\sin(1/x^4)$. Then $f$ is continuous, but $f'$ is not. – David C. Ullrich Aug 23 '22 at 10:32
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    I can't answer the question of what's wrong with the proof you give because I cannot tell exactly what you're trying to prove. You say throughout the proof that if we assume this then one thing while if we assume something else then something else, but at the end I can't tell exactly what you are assuming – David C. Ullrich Aug 23 '22 at 10:35
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    @DavidC.Ullrich There are two assumptions. $f$ is continuous at $a$. And $\displaystyle \lim_{x \to a^+}f'(x)$ exists. – S.C. Aug 23 '22 at 10:36

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